The binomial theorem

• Use Pascal’s Triangle to expand a binomial. (IA 12.4.1)

Pascal’s triangle helps us find the coefficients of the terms in the expansion of a binomial.

To find the coefficients of the terms, we write our expansion again focusing on the coefficients. We rewrite the coefficients to the right forming an array of coefficients. The array to the right is called Pascal’s Triangle.

Notice that in each expansion the powers of a in each term decrease from n to 0, and the powers of b increase from 0 to n.

Notice each number in the array is the sum of the two closest numbers in the row above. We can find the next row by starting and ending with one and then adding two adjacent numbers.

To find the coefficients of the expansion of the binomial \({(a+b)}^{n}\) , go to the row that has the value n as a second entry.

Use Pascal’s Triangle to expand a binomial.

In Counting Principles, we studied combinations. In the shortcut to finding \({(x+y)}^{n},\) we will need to use combinations to find the coefficients that will appear in the expansion of the binomial. In this case, we use the notation \((\begin{array}{l}n \\ r\end{array})\) instead of \(C(n,r),\) but it can be calculated in the same way. So\[(\begin{array}{l}n \\ r\end{array})=C(n,r)=\frac{n!}{r!(n-r)!}\]

The combination \((\begin{array}{l}n \\ r\end{array})\) is called a binomial coefficient. An example of a binomial coefficient is \((\begin{array}{l}5 \\ 2\end{array})=C(5,2)=10.\)

When we expand \({(x+y)}^{n}\) by multiplying, the result is called a binomial expansion, and it includes binomial coefficients. If we wanted to expand \({(x+y)}^{52},\) we might multiply \((x+y)\) by itself fifty-two times. This could take hours! If we examine some simple binomial expansions, we can find patterns that will lead us to a shortcut for finding more complicated binomial expansions.

First, let’s examine the exponents. With each successive term, the exponent for \(x\) decreases and the exponent for \(y\) increases. The sum of the two exponents is \(n\) for each term.

Next, let’s examine the coefficients. Notice that the coefficients increase and then decrease in a symmetrical pattern. The coefficients follow a pattern:

These patterns lead us to the Binomial Theorem, which can be used to expand any binomial.

Another way to see the coefficients is to examine the expansion of a binomial in general form, \(x+y,\) to successive powers 1, 2, 3, and 4.

Can you guess the next expansion for the binomial \({(x+y)}^{5}?\)

See , which illustrates the following:

• There are \(n+1\) terms in the expansion of \({(x+y)}^{n}.\)
• The degree (or sum of the exponents) for each term is \(n.\)
• The powers on \(x\) begin with \(n\) and decrease to 0.
• The powers on \(y\) begin with 0 and increase to \(n.\)
• The coefficients are symmetric.

• Introduce \({x}^{5},\) and then for each successive term reduce the exponent on \(x\) by 1 until \({x}^{0}=1\) is reached.
• Introduce \({y}^{0}=1,\) and then increase the exponent on \(y\) by 1 until \({y}^{5}\) is reached. \[{x}^{5},\ {x}^{4}y,\ {x}^{3}{y}^{2},\ {x}^{2}{y}^{3},\ x{y}^{4},\ {y}^{5}\]

Condensed: the full section is in OpenStax College Algebra 2e.

Expanding a binomial with a high exponent such as \({(x+2y)}^{16}\) can be a lengthy process.

Sometimes we are interested only in a certain term of a binomial expansion. We do not need to fully expand a binomial to find a single specific term.

Note the pattern of coefficients in the expansion of \({(x+y)}^{5}.\)

The second term is \((\begin{array}{l}5 \\ 1\end{array}){x}^{4}y.\) The third term is \((\begin{array}{l}5 \\ 2\end{array}){x}^{3}{y}^{2}.\) We can generalize this result.

• \((\begin{array}{l}n \\ r\end{array})\) is called a binomial coefficient and is equal to \(C(n,r).\)
• The Binomial Theorem allows us to expand binomials without multiplying.
• We can find a given term of a binomial expansion without fully expanding the binomial.

• Use Pascal’s Triangle to expand a binomial. (IA 12.4.1)

Pascal’s triangle helps us find the coefficients of the terms in the expansion of a binomial.

To find the coefficients of the terms, we write our expansion again focusing on the coefficients. We rewrite the coefficients to the right forming an array of coefficients. The array to the right is called Pascal’s Triangle.

Notice that in each expansion the powers of a in each term decrease from n to 0, and the powers of b increase from 0 to n.

Notice each number in the array is the sum of the two closest numbers in the row above. We can find the next row by starting and ending with one and then adding two adjacent numbers.

To find the coefficients of the expansion of the binomial \({(a+b)}^{n}\) , go to the row that has the value n as a second entry.

Use Pascal’s Triangle to expand a binomial.

In Counting Principles, we studied combinations. In the shortcut to finding \({(x+y)}^{n},\) we will need to use combinations to find the coefficients that will appear in the expansion of the binomial. In this case, we use the notation \((\begin{array}{l}n \\ r\end{array})\) instead of \(C(n,r),\) but it can be calculated in the same way. So\[(\begin{array}{l}n \\ r\end{array})=C(n,r)=\frac{n!}{r!(n-r)!}\]

The combination \((\begin{array}{l}n \\ r\end{array})\) is called a binomial coefficient. An example of a binomial coefficient is \((\begin{array}{l}5 \\ 2\end{array})=C(5,2)=10.\)

When we expand \({(x+y)}^{n}\) by multiplying, the result is called a binomial expansion, and it includes binomial coefficients. If we wanted to expand \({(x+y)}^{52},\) we might multiply \((x+y)\) by itself fifty-two times. This could take hours! If we examine some simple binomial expansions, we can find patterns that will lead us to a shortcut for finding more complicated binomial expansions.

First, let’s examine the exponents. With each successive term, the exponent for \(x\) decreases and the exponent for \(y\) increases. The sum of the two exponents is \(n\) for each term.

Next, let’s examine the coefficients. Notice that the coefficients increase and then decrease in a symmetrical pattern. The coefficients follow a pattern:

These patterns lead us to the Binomial Theorem, which can be used to expand any binomial.

Another way to see the coefficients is to examine the expansion of a binomial in general form, \(x+y,\) to successive powers 1, 2, 3, and 4.

Can you guess the next expansion for the binomial \({(x+y)}^{5}?\)

See , which illustrates the following:

• There are \(n+1\) terms in the expansion of \({(x+y)}^{n}.\)
• The degree (or sum of the exponents) for each term is \(n.\)
• The powers on \(x\) begin with \(n\) and decrease to 0.
• The powers on \(y\) begin with 0 and increase to \(n.\)
• The coefficients are symmetric.

• Introduce \({x}^{5},\) and then for each successive term reduce the exponent on \(x\) by 1 until \({x}^{0}=1\) is reached.
• Introduce \({y}^{0}=1,\) and then increase the exponent on \(y\) by 1 until \({y}^{5}\) is reached. \[{x}^{5},\ {x}^{4}y,\ {x}^{3}{y}^{2},\ {x}^{2}{y}^{3},\ x{y}^{4},\ {y}^{5}\]

Condensed: the full section is in OpenStax Precalculus 2e.

Expanding a binomial with a high exponent such as \({(x+2y)}^{16}\) can be a lengthy process.

Sometimes we are interested only in a certain term of a binomial expansion. We do not need to fully expand a binomial to find a single specific term.

Note the pattern of coefficients in the expansion of \({(x+y)}^{5}.\)

The second term is \((\begin{array}{l}5 \\ 1\end{array}){x}^{4}y.\) The third term is \((\begin{array}{l}5 \\ 2\end{array}){x}^{3}{y}^{2}.\) We can generalize this result.

• \((\begin{array}{l}n \\ r\end{array})\) is called a binomial coefficient and is equal to \(C(n,r).\)
• The Binomial Theorem allows us to expand binomials without multiplying.
• We can find a given term of a binomial expansion without fully expanding the binomial.

In our previous work, we have squared binomials either by using FOIL or by using the Binomial Squares Pattern. We can also say that we expanded \({(a+b)}^{2}.\)

To expand \({(a+b)}^{3},\) we recognize that this is \({(a+b)}^{2}(a+b)\) and multiply.

To find a method that is less tedious that will work for higher expansions like \({(a+b)}^{7},\) we again look for patterns in some expansions.

Notice the first and last terms show only one variable. Recall that \({a}^{0}=1,\) so we could rewrite the first and last terms to include both variables. For example, we could expand \({(a+b)}^{3}\) to show each term with both variables.

Generally, we don’t show the zero exponents, just as we usually write x rather than 1x.

Let’s look at an example to highlight the last three patterns.

From the patterns we identified, we see the variables in the expansion of \({(a+b)}^{n},\) would be

Condensed: the full section is in OpenStax Intermediate Algebra 2e.

While Pascal’s Triangle is one method to expand a binomial, we will also look at another method. Before we get to that, we need to introduce some more factorial notation. This notation is not only used to expand binomials, but also in the study and use of probability.

To find the coefficients of the terms of expanded binomials, we will need to be able to evaluate the notation \((\begin{array}{l}n \\ r\end{array})\) which is called a binomial coefficient. We read \((\begin{array}{l}n \\ r\end{array})\) as “n choose r” or “n taken r at a time”.

Condensed: the full section is in OpenStax Intermediate Algebra 2e.

We are now ready to use the alternate method of expanding binomials. The Binomial Theorem uses the same pattern for the variables, but uses the binomial coefficient for the coefficient of each term.

Notice that when we expanded \({(p+q)}^{4}\) in the last example, using the Binomial Theorem, we got the same coefficients we would get from using Pascal’s Triangle.

The next example, the binomial is a difference. When the binomial is a difference, we must be careful in identifying the values we will use in the pattern.

Condensed: the full section is in OpenStax Intermediate Algebra 2e.

• Patterns in the expansion of \({(a+b)}^{n}\)
  • The number of terms is \(n+1.\)
  • The first term is \({a}^{n}\) and the last term is \({b}^{n}.\)
  • The exponents on a decrease by one on each term going left to right.
  • The exponents on b increase by one on each term going left to right.
  • The sum of the exponents on any term is n.
• Pascal’s Triangle
• Binomial Coefficient \((\begin{array}{l}n \\ r\end{array})\) : A binomial coefficient \((\begin{array}{l}n \\ r\end{array}),\) where r and n are integers with \(0\le r\le n,\) is defined as
  \[(\begin{array}{l}n \\ r\end{array})=\frac{n!}{r!(n-r)!}\]
  We read \((\begin{array}{l}n \\ r\end{array})\) as “n choose r” or “n taken r at a time”.
• Properties of Binomial Coefficients
  \(\begin{array}{lllllllll}(\begin{array}{l}n \\ 1\end{array})=n & & & & (\begin{array}{l}n \\ n\end{array})=1 & & & & (\begin{array}{l}n \\ 0\end{array})=1\end{array}\)
• Binomial Theorem: For any real numbers a, b, and positive integer n,
  \[{(a+b)}^{n}=(\begin{array}{l}n \\ 0\end{array}){a}^{n}+(\begin{array}{l}n \\ 1\end{array}){a}^{n-1}{b}^{1}+(\begin{array}{l}n \\ 2\end{array}){a}^{n-2}{b}^{2}+...+(\begin{array}{l}n \\ r\end{array}){a}^{n-r}{b}^{r}+...+(\begin{array}{l}n \\ n\end{array}){b}^{n}\]

In elementary algebra, the binomial theorem (or binomial expansion) describes the algebraic expansion of powers of a binomial. According to the theorem, the power ⁠\(\textstyle (x+y)^n\)⁠ expands into a polynomial with terms of the form ⁠\(\textstyle ax^ky^m\)⁠, where the exponents ⁠\(k\)⁠ and ⁠\(m\)⁠ are nonnegative integers satisfying ⁠\(k + m = n\)⁠ and the coefficient ⁠\(a\)⁠ of each term is a specific positive integer depending on ⁠\(n\)⁠ and ⁠\(k\)⁠. For example, for ⁠\(n = 4\)⁠, \[(x+y)^4 = x^4 + 4 x^3y + 6 x^2 y^2 + 4 x y^3 + y^4.\]

The coefficient ⁠\(a\)⁠ in each term ⁠\(\textstyle ax^ky^m\)⁠ is known as the binomial coefficient ⁠\(\tbinom nk\)⁠ or ⁠\(\tbinom{n}{m}\)⁠ (the two have the same value). These coefficients for varying ⁠\(n\)⁠ and ⁠\(k\)⁠ can be arranged to form Pascal's triangle. These numbers also occur in combinatorics, where ⁠\(\tbinom nk\)⁠ gives the number of different combinations (i.e. subsets) of ⁠\(k\)⁠ elements that can be chosen from an ⁠\(n\)⁠-element set. Therefore ⁠\(\tbinom nk\)⁠ is usually pronounced as "⁠\(n\)⁠ choose ⁠\(k\)⁠".

According to the theorem, the expansion of any nonnegative integer power n of the binomial x + y is a sum of the form \[(x+y)^n = {\binom{n}{0}}x^n y^0 + {\binom{n}{1}}x^{n-1} y^1 + {\binom{n}{2}}x^{n-2} y^2 + \cdots + {\binom{n}{n}}x^0 y^n,\] where each \(\tbinom nk\) is a positive integer known as a binomial coefficient, defined as

\[\binom nk = \frac{n!}{k!\,(n-k)!} = \frac{n(n-1)(n-2)\cdots(n-k + 1)}{k(k-1)(k-2)\cdots2\cdot1}.\]

This formula is also referred to as the binomial formula or the binomial identity. Using summation notation, it can be written more concisely as \[(x+y)^n = \sum_{k=0}^n {\binom{n}{k}}x^{n-k}y^k = \sum_{k=0}^n {\binom{n}{k}}x^{k}y^{n-k}.\]

The final expression follows from the previous one by the symmetry of x and y in the first expression, and by comparison it follows that the sequence of binomial coefficients in the formula is symmetric, \(\binom nk = \binom n{n-k}.\)

A simple variant of the binomial formula is obtained by substituting 1 for y, so that it involves only a single variable. In this form, the formula reads \[\begin{aligned} (x+1)^n &= {\binom{n}{0}}x^0 + {\binom{n}{1}}x^1 + {\binom{n}{2}}x^2 + \cdots + {\binom{n}{n}}x^n \\[4mu] &= \sum_{k=0}^n {\binom{n}{k}}x^k. \vphantom{\Bigg)} \end{aligned}\]

The first few cases of the binomial theorem are: \[\begin{aligned} (x+y)^0 & = 1, \\[2mu] (x+y)^1 & = x + y, \\[2mu] (x+y)^2 & = x^2 + 2xy + y^2, \\[2mu] (x+y)^3 & = x^3 + 3x^2y + 3xy^2 + y^3, \\[2mu] (x+y)^4 & = x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4, \end{aligned}\] In general, for the expansion of (x + y) on the right side in the nth row (numbered so that the top row is the 0th row):

• the exponents of x in the terms are n, n − 1, ..., 2, 1, 0 (the last term implicitly contains x = 1);
• the exponents of y in the terms are 0, 1, 2, ..., n − 1, n (the first term implicitly contains y = 1);
• the coefficients form the nth row of Pascal's triangle;
• before combining like terms, there are 2 terms xy in the expansion (not shown);
• after combining like terms, there are n + 1 terms, and their coefficients sum to 2.

An example illustrating the last two points: \[\begin{aligned} (x+y)^3 & = xxx + xxy + xyx + xyy + yxx + yxy + yyx + yyy & (2^3 \text{ terms}) \\ & = x^3 + 3x^2y + 3xy^2 + y^3 & (3 + 1 \text{ terms}) \end{aligned}\] with \(1 + 3 + 3 + 1 = 2^3\).

A simple example with a specific positive value of y: \[\begin{aligned} (x+2)^3 &= x^3 + 3x^2(2) + 3x(2)^2 + 2^3 \\ &= x^3 + 6x^2 + 12x + 8. \end{aligned}\]

A simple example with a specific negative value of y: \[\begin{aligned} (x-2)^3 &= x^3 - 3x^2(2) + 3x(2)^2 - 2^3 \\ &= x^3 - 6x^2 + 12x - 8. \end{aligned}\]

For positive values of a and b, the binomial theorem with n = 2 is the geometrically evident fact that a square of side a + b can be cut into a square of side a, a square of side b, and two rectangles with sides a and b. With n = 3, the theorem states that a cube of side a + b can be cut into a cube of side a, a cube of side b, three a × a × b rectangular boxes, and three a × b × b rectangular boxes.

In calculus, this picture also gives a geometric proof of the derivative \((x^n)'=nx^{n-1}:\) if one sets \(a=x\) and \(b=\Delta x,\) interpreting b as an infinitesimal change in a, then this picture shows the infinitesimal change in the volume of an n-dimensional hypercube, \((x+\Delta x)^n,\) where the coefficient of the linear term (in \(\Delta x\)) is \(nx^{n-1},\) the area of the n faces, each of dimension n − 1: \[(x+\Delta x)^n = x^n + nx^{n-1}\Delta x + \binom{n}{2}x^{n-2}(\Delta x)^2 + \cdots.\] Substituting this into the definition of the derivative via a difference quotient and taking limits means that the higher order terms, \((\Delta x)^2\) and higher, become negligible, and yields the formula \((x^n)'=nx^{n-1},\) interpreted as "the infinitesimal rate of change in volume of an n-cube as side length varies is the area of n of its (n − 1)-dimensional faces". If one integrates this picture, which corresponds to applying the fundamental theorem of calculus, one obtains Cavalieri's quadrature formula, the integral \(\textstyle{\int x^{n-1}\,dx = \tfrac{1}{n} x^n}\), see proof of Cavalieri's quadrature formula for details.

The coefficients that appear in the binomial expansion are called binomial coefficients. These are usually written \(\tbinom{n}{k},\) and pronounced "n choose k".

The coefficient of xy is given by the formula \[\binom{n}{k} = \frac{n!}{k! \; (n-k)!},\] which is defined in terms of the factorial function n!. Equivalently, this formula can be written \[\binom{n}{k} = \frac{n (n-1) \cdots (n-k+1)}{k (k-1) \cdots 1} = \prod_{\ell=1}^k \frac{n-\ell+1}{\ell} = \prod_{\ell=0}^{k-1} \frac{n-\ell}{k - \ell}\] with k factors in both the numerator and denominator of the fraction. Although this formula involves a fraction, the binomial coefficient \(\tbinom{n}{k}\) is actually an integer.

The binomial coefficient \(\tbinom nk\) can be interpreted as the number of ways to choose k elements from an n-element set (a combination). This is related to binomials for the following reason: if we write (x + y) as a product \[(x+y)(x+y)(x+y)\cdots(x+y),\] then, according to the distributive law, there will be one term in the expansion for each choice of either x or y from each of the binomials of the product. For example, there will only be one term x, corresponding to choosing x from each binomial. However, there will be several terms of the form xy, one for each way of choosing exactly two binomials to contribute a y. Therefore, after combining like terms, the coefficient of xy will be equal to the number of ways to choose exactly 2 elements from an n-element set.

Expanding (x + y) yields the sum of the 2 products of the form e₁e₂ ... e_n where each e_i is x or y. Rearranging factors shows that each product equals xy for some k between 0 and n. For a given k, the following are proved equal in succession:

• the number of terms equal to xy in the expansion
• the number of n-character x,y strings having y in exactly k positions
• the number of k-element subsets of {1, 2, ..., n}
• \(\tbinom{n}{k},\) either by definition, or by a short combinatorial argument if one is defining \(\tbinom{n}{k}\) as \(\tfrac{n!}{k! (n-k)!}.\)

This proves the binomial theorem.

Induction yields another proof of the binomial theorem. When n = 0, both sides equal 1, since x = 1 and \(\tbinom{0}{0}=1.\) Now suppose that the equality holds for a given n; we will prove it for n + 1. For j, k ≥ 0, let [f(x, y)]_j,k denote the coefficient of xy in the polynomial f(x, y). By the inductive hypothesis, (x + y) is a polynomial in x and y such that [(x + y)]_j,k is \(\tbinom{n}{k}\) if j + k = n, and 0 otherwise. The identity \[(x+y)^{n+1} = x(x+y)^n + y(x+y)^n\] shows that (x + y) is also a polynomial in x and y, and \[[(x+y)^{n+1}]_{j,k} = [(x+y)^n]_{j-1,k} + [(x+y)^n]_{j,k-1},\] since if j + k = n + 1, then (j − 1) + k = n and j + (k − 1) = n. Now, the right hand side is \[\binom{n}{k} + \binom{n}{k-1} = \binom{n+1}{k},\] by Pascal's identity. On the other hand, if j + k ≠ n + 1, then (j, 1) + k ≠ n and j + (k, 1) ≠ n, so we get 0 + 0 = 0. Thus \[(x+y)^{n+1} = \sum_{k=0}^{n+1} \binom{n+1}{k} x^{n+1-k} y^k,\] which is the inductive hypothesis with n + 1 substituted for n and so completes the inductive step.

The standard binomial theorem, as discussed above, is concerned with \((x+y)^n\) where the exponent \(n\) is a nonnegative integer. The generalized binomial theorem allows for non-integer, negative, or even complex exponents, at the expense of replacing the finite sum by an infinite series.

In order to do this, one needs to give meaning to binomial coefficients with an arbitrary upper index, which cannot be done using the usual formula with factorials. However, for an arbitrary number r, one can define \[{\binom{r}{k}}=\frac{r(r-1) \cdots (r-k+1)}{k!} =\frac{r^{\underline{k}}}{k!},\] where the last equation introduces modern notation for the falling factorial. This agrees with the usual definitions when r is a nonnegative integer. Then, if x and y are real numbers with |x| > |y|, and r is any complex number, one has \[\begin{aligned} (x+y)^r & =\sum_{k=0}^\infty {\binom{r}{k}} x^{r-k} y^k \\ &= x^r + r x^{r-1} y + \frac{r(r-1)}{2!} x^{r-2} y^2 + \frac{r(r-1)(r-2)}{3!} x^{r-3} y^3 + \cdots. \end{aligned}\]

When r is a nonnegative integer, the binomial coefficients for k > r are zero, so this equation reduces to the usual binomial theorem, and there are at most r + 1 nonzero terms. For other values of r, the series has infinitely many nonzero terms.

For example, r = 1/2 gives the following series for the square root: \[\sqrt{1+x} = 1 + \frac{1}{2}x - \frac{1}{8}x^2 + \frac{1}{16}x^3 - \frac{5}{128}x^4 + \frac{7}{256}x^5 - \cdots.\]

With r = −1, the generalized binomial series becomes: \[(1+x)^{-1} = \frac{1}{1+x} = 1 - x + x^2 - x^3 + x^4 - x^5 + \cdots.\] which is the geometric series sum formula for the convergent case |x| < 1, whose common ratio is −x.

More generally, with r = −s, we have for |x| < 1: \[\frac{1}{(1+x)^s} = \sum_{k=0}^\infty {\binom{-s}{k}} x^k = \sum_{k=0}^\infty {\binom{s+k-1}{k}} (-1)^k x^k.\]

So, for instance, when s = 1/2, \[\frac{1}{\sqrt{1+x}} = 1 - \frac{1}{2}x + \frac{3}{8}x^2 - \frac{5}{16}x^3 + \frac{35}{128}x^4 - \frac{63}{256}x^5 + \cdots.\]

Condensed: the full section is in Wikipedia.

The generalized binomial theorem can be extended to the case where x and y are complex numbers. For this version, one should again assume |x| > |y| and define the powers of x + y and x using a holomorphic branch of log defined on an open disk of radius |x| centered at x. The generalized binomial theorem is valid also for elements x and y of a Banach algebra as long as xy = yx, and x is invertible, and ‖y/x‖ < 1.

A version of the binomial theorem is valid for the following Pochhammer symbol-like family of polynomials: for a given real constant c, define \(x^{(0)} = 1\) and \[x^{(n)} = \prod_{k=1}^{n}[x+(k-1)c]\] for \(n > 0.\) Then \[(a + b)^{(n)} = \sum_{k=0}^{n}\binom{n}{k}a^{(n-k)}b^{(k)}.\] The case c = 0 recovers the usual binomial theorem.

More generally, a sequence \(\{p_n\}_{n=0}^\infty\) of polynomials is said to be of binomial type if

• \(\deg p_n = n\) for all \(n\),
• \(p_0(0) = 1\), and
• \(p_n(x+y) = \sum_{k=0}^n \binom{n}{k} p_k(x) p_{n-k}(y)\) for all \(x\), \(y\), and \(n\).

An operator \(Q\) on the space of polynomials is said to be the basis operator of the sequence \(\{p_n\}_{n=0}^\infty\) if \(Qp_0 = 0\) and \(Q p_n = n p_{n-1}\) for all \(n \geqslant 1\). A sequence \(\{p_n\}_{n=0}^\infty\) is binomial if and only if its basis operator is a Delta operator. Writing \(E^a\) for the shift by \(a\) operator, the Delta operators corresponding to the above "Pochhammer" families of polynomials are the backward difference \(I - E^{-c}\) for \(c>0\), the ordinary derivative for \(c=0\), and the forward difference \(E^{-c} - I\) for \(c<0\).

The binomial theorem can be generalized to include powers of sums with more than two terms. The general version is

\[(x_1 + x_2 + \cdots + x_m)^n = \sum_{k_1+k_2+\cdots +k_m = n} \binom{n}{k_1, k_2, \ldots, k_m} x_1^{k_1} x_2^{k_2} \cdots x_m^{k_m},\]

where the summation is taken over all sequences of nonnegative integer indices k₁ through k_m such that the sum of all k_i is n. (For each term in the expansion, the exponents must add up to n). The coefficients \(\tbinom{n}{k_1,\cdots,k_m}\) are known as multinomial coefficients, and can be computed by the formula \[\binom{n}{k_1, k_2, \ldots, k_m} = \frac{n!}{k_1! \cdot k_2! \cdots k_m!}.\]

Combinatorially, the multinomial coefficient \(\tbinom{n}{k_1,\cdots,k_m}\) counts the number of different ways to partition an n-element set into disjoint subsets of sizes k₁, ..., k_m.

When working in more dimensions, it is often useful to deal with products of binomial expressions. By the binomial theorem this is equal to \[(x_1+y_1)^{n_1}\dotsm(x_d+y_d)^{n_d} = \sum_{k_1=0}^{n_1}\dotsm\sum_{k_d=0}^{n_d} \binom{n_1}{k_1} x_1^{k_1}y_1^{n_1-k_1} \dotsc \binom{n_d}{k_d} x_d^{k_d}y_d^{n_d-k_d}.\]

This may be written more concisely, by multi-index notation, as \[(x+y)^\alpha = \sum_{\nu \le \alpha} \binom{\alpha}{\nu} x^\nu y^{\alpha - \nu}.\]

The general Leibniz rule gives the nth derivative of a product of two functions in a form similar to that of the binomial theorem: \[(fg)^{(n)}(x) = \sum_{k=0}^n \binom{n}{k} f^{(n-k)}(x) g^{(k)}(x).\]

Here, the superscript (n) indicates the nth derivative of a function, \(f^{(n)}(x) = \tfrac{d^n}{dx^n}f(x)\). If one sets f(x) = e and g(x) = e, cancelling the common factor of e from each term gives the ordinary binomial theorem.